How do I find the points of intersection; x= y^2 + 3 and y= x-5?

Question

I know I need to set it equal to y; 3-x = (x-5)^2
I am not getting the right answer...

Answer 1

If y = x-5, then x = y+5, so solve y+5 = y^2+3 by the quadratic formula. Then substitute in either equation to get the corresponding x values.

Answer 2

The first expression actually gives you x-3 = y^2, not 3-x = y^2. But if you use the same strategy you get:

(x-3) = (x-5)^2
x - 3 = x^2 - 10x + 25
0 = x^2 - 11x + 28

This can easily be factored into (x-4)(x-7)=0, so x=4 or 7. Plug this into y=x-5 and you get -1 and 2. So the points of intersection are (4,-1) and (7,2).

Answer 3

First, take the equation x = y^2+3 and substitute x-5 in for y
You get x = (x-5)^2+3 Now simplify to
x = x^2-10x+25+3
Get this equation equal to zero
x^2-11x+28 = 0
Now factor to
(x-7)(x-4) = 0 and solve for x
x = 7 or x =4

Now plug those values back into either of the original equations to get the y value (I suggest using y = x - 5)

Now you have the points (7, 2) and (4, -1) and these are your answers.

You might notice that using x = y^2+3 will give you two possible answers for each value of x, however only one answer for each x will apply to the second equation, so make sure that the answer you use works for both of them.

Answer 4

Solve the second equation for x, x = y+5, and plug it in to the first equation: y+5 = y² + 3. That's a solvable quadratic.

y² + 3 - y - 5 = 0
y² - y - 2 = 0
(y-2)(y+1) = 0
y = 2, x = 7 or
y = -1, x = 4,

so intersections at (7,2) and (4,-1)

Answer 5

This one is easy, you just have to equal both equations, so let's try it:

x=y^2+3

y=x-5, you solve for x so x=y+5

so,

y+5=y^2+3

Then you solve for y

y^2-y-2=0

(y-2)(y+1)=0

y= 2 , -1

Then you plug those values in either of the two equations...

2=x-5

x=7 Meaning (7,2) is a point of intersection.

and you plug the second value

-1=x-5

x=4 Meaning (4,-1) is another point of intersection.

That's it, you will notice there are two points of intersection since one equation is a parabola, and the other is a line with a slope, so the line will cross two times in the parabolas path.

Answer 6

equalize 2 equation to each other.

y+5=y^2+3 Therefore;
y^2-y-2=0
y=2 or y=-1

use this y s in above equalities

points are (7,2) and (4,-1)

Answer 7

solve both for y
y=x-5
y=sqrt(x-3)

set them equal to one another
x-5=sqrt(x-3

square both sides
x^2 - 10x +25= x-3

make equation equal to 0
x^2 - 11x +28

use quadratic formula for intersection points
x= 2/7 or 1/2

Answer 8

x-3=(x-5)^2 thats why you are not getting right answer